Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $r = \dfrac{6q + 42}{-10q - 40} \times \dfrac{q + 1}{q^2 + 8q + 7} $
Solution: First factor the quadratic. $r = \dfrac{6q + 42}{-10q - 40} \times \dfrac{q + 1}{(q + 7)(q + 1)} $ Then factor out any other terms. $r = \dfrac{6(q + 7)}{-10(q + 4)} \times \dfrac{q + 1}{(q + 7)(q + 1)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ 6(q + 7) \times (q + 1) } { -10(q + 4) \times (q + 7)(q + 1) } $ $r = \dfrac{ 6(q + 7)(q + 1)}{ -10(q + 4)(q + 7)(q + 1)} $ Notice that $(q + 1)$ and $(q + 7)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ 6\cancel{(q + 7)}(q + 1)}{ -10(q + 4)\cancel{(q + 7)}(q + 1)} $ We are dividing by $q + 7$ , so $q + 7 \neq 0$ Therefore, $q \neq -7$ $r = \dfrac{ 6\cancel{(q + 7)}\cancel{(q + 1)}}{ -10(q + 4)\cancel{(q + 7)}\cancel{(q + 1)}} $ We are dividing by $q + 1$ , so $q + 1 \neq 0$ Therefore, $q \neq -1$ $r = \dfrac{6}{-10(q + 4)} $ $r = \dfrac{-3}{5(q + 4)} ; \space q \neq -7 ; \space q \neq -1 $